python - Comment demander une nouvelle URL à l'aide de Scrapy et rappeler la fonction spécifiée?

Question

À propos de l'utilisation de Scrapy sous Python3

import re
import scrapy
from bs4 import BeautifulSoup
from scrapy.http import Request
from ..items import ZhibobaItem
import json
import lxml.html
import requests
import json


class Myspider(scrapy.Spider):
    name = 'zhiboba'
    allowed_domains = ['zhibo8.cc']
    json_url = 'https://bifen4pc.qiumibao.com/json/list.htm?85591'
    bash_url = 'https://www.zhibo8.cc/'

    def start_requests(self):
        yield Request(self.bash_url, self.parse_index)

    def parse_index(self, response):
        print("enter the parse_index")
        print(self.bash_url)
        ps = BeautifulSoup(response.text, 'lxml').find_all(label=re.compile("足球"))
        item = ZhibobaItem()
        for single_p in ps:
            item['label'] = single_p.get('label')
            item['sdate'] = single_p.get('data-time')
            item['linkurl'] = self.bash_url + single_p.find('a')['href']
            home_team = single_p.get_text().split()[2]
            item['home_team'] = home_team
            visit_team = single_p.get_text().split()[4]
            item['visit_team'] = visit_team
            print("quit the parse_index")
            print(self.json_url)
            yield Request(self.json_url, callback=self.get_score, meta={'home_team': home_team,
                                                                        'visit_team': visit_team
                                                                        })
    def get_score(self, response):
        print("enter the get_score")
        json_url = self.json_url
        wbdata = response.get(json_url).text
        data = json.loads(wbdata)
        news = data['list']
        print(wbdata)
        print("quit the get_score")

Lorsque j'exécute le code ci-dessus, je ne parviens pas à appeler json_url et la fonction de réponse correspondante get_score.

Answer 1

Essayez de modifierallow_domains = [].